` int _(0)^(pi) cos^(3) x dx` is equal to

Options

A. 0
B. 1
C. `-1`
D. `(1)/(2sqrt2)`

Explanation

To solve the integral \( I = \int_0^{\pi} \cos^3 x \, dx \), we will use a trigonometric identity to simplify the integrand.

### Step-by-Step Solution:

1. **Set Up the Integral:**
\[ I = \int_0^{\pi} \cos^3 x \, dx \]

2. **Use the Trigonometric Identity:**
We know that:
\[ \cos 3x = 4 \cos^3 x - 3 \cos x \]
Rearranging this gives:
\[ \cos^3 x = \frac{1}{4} (\cos 3x + 3 \cos x) \]

3. **Substitute the Identity into the Integral:**
Substitute the expression for \( \cos^3 x \) into the integral:
\[ I = \int_0^{\pi} \frac{1}{4} (\cos 3x + 3 \cos x) \, dx \]
This simplifies to:
\[ I = \frac{1}{4} \left( \int_0^{\pi} \cos 3x \, dx + 3 \int_0^{\pi} \cos x \, dx \right) \]

4. **Evaluate the Integrals:**
- First, evaluate \( \int_0^{\pi} \cos x \, dx \):
\[ \int_0^{\pi} \cos x \, dx = [\sin x]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 \]

- Next, evaluate \( \int_0^{\pi} \cos 3x \, dx \):
\[ \int_0^{\pi} \cos 3x \, dx = \left[ \frac{1}{3} \sin 3x \right]_0^{\pi} = \frac{1}{3} (\sin(3\pi) - \sin(0)) = \frac{1}{3} (0 - 0) = 0 \]

5. **Combine the Results:**
Substitute the results of the integrals back into the equation for \( I \):
\[ I = \frac{1}{4} \left( 0 + 3 \cdot 0 \right) = \frac{1}{4} \cdot 0 = 0 \]

6. **Final Result:**
Therefore, the value of the integral is:
\[ I = 0 \]